Question

The planes $bx-ay=n,cy-bz=1,az-cx=m$ intersect in a line if

• A. $al+bm+cn=0$
• B. $al-bm+cn=0$
• C. $al-bm-cn+1=0$
• D. $al+bm+cn=1$

Answer 1

The correct option is A $al+bm+cn=0$

Since the planes

$bx-ay=n$, $\Rightarrow x=\frac{n+ay}{b}⟶\left(1\right)$

$cy-bz=l$, $\Rightarrow z=\frac{cy-l}{b}⟶\left(2\right)$

and, $az-cx=m$

intersect in a line, eliminating $x,y,z$ we will get the desired condition.

Substitute $\left(1\right)$ and $\left(2\right)$ in $\left(3\right)$

$a\left(\frac{cy-l}{b}\right)-c\left(\frac{n+ay}{b}\right)=m$

$\Rightarrow acy-al-cn-acy=mb$

$\Rightarrow al+cn+bm=0$

$\Rightarrow al+bm+cn=0$

Answer : Option B.